`1,` Ta có:
`1/4\sqrt{82}=\sqrt{1/16. 82}=\sqrt{41/8}=\sqrt{5 1/8}`
`6\sqrt{1/7}=\sqrt{36. 1/7}=\sqrt{36/7}=\sqrt{5 1/7}`
Do `1/7>1/8`
`<=> 5 1/7>5 1/8`
`<=> 6\sqrt{1/7}>1/4\sqrt{82}`
Vậy `6\sqrt{1/7}>1/4\sqrt{82}`
`2, A=6\sqrt{2}+\sqrt{6-\sqrt{11}}-\sqrt{6+\sqrt{11}}`
`A=(12+\sqrt{12-2\sqrt{11}}-\sqrt{12+2\sqrt{11}})/\sqrt{2}`
`A=(12+sqrt{11-2\sqrt{11}+1}-\sqrt{11+2sqrt{11}+1})/\sqrt{2}`
`A=(12+sqrt{(sqrt{11}-1)^2}-sqrt{(sqrt{11}+1)^2})/\sqrt{2}`
`A=(12+|sqrt{11}-1|-|sqrt{11}+1|)/\sqrt{2}`
`A=(12+sqrt{11}-1-sqrt{11}-1)/\sqrt{2}`
`A=(12-2)/\sqrt{2}`
`A=10/\sqrt{2}`
`A=5sqrt{2}`
